Answer:
The 95% confidence interval is [tex]98.27 < \mu < 102.53[/tex]
This interval means that there 95% confidence that the true mean is within this interval
Yes i would agree with my friend because the lower and the upper limit 95% confidence interval for mean points scored at home is greater than 98 points
Step-by-step explanation:
From the question we are told that
The sample size is n = 20
The sample mean is [tex]\mu = 100.4[/tex]
The standard deviation is [tex]\sigma = 4.86[/tex]
Given that the confidence level is 95% then the level of significance is mathematically evaluated as
[tex]\alpha = 100 - 95[/tex]
[tex]\alpha = 5\%[/tex]
[tex]\alpha = 0.05[/tex]
Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table, the value is
[tex]Z_{\frac{ \alpha }{2} } = 1.96[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{ \sqrt{n} }[/tex]
substituting values
[tex]E = 1.96 * \frac{ 4.86 }{ \sqrt{20 } }[/tex]
[tex]E = 2.13[/tex]
The 95% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
substituting values
[tex]100.4 - 2.13 < \mu < 100.4 + 2.13[/tex]
[tex]98.27 < \mu < 102.53[/tex]
